An arrow in flight has an initial velocity of 65 meters per second, and 10 seconds later, it has a velocity of 35 meters per sec
1 answer:
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Answer:
17.66 kPa
Explanation:
The volume of water in the swimming pool is the product of its dimensions
V = 30 * 8.7 * 1.8 = 469.8 cubic meters
Let water density , and g = 9.81 m/s2 we can calculate the total weight of water in the swimming pool
The area of the bottom
A = 30 * 8.7 = 261 square meters
Therefore the pressure is its force over unit area
or 17.66 kPa
We can find the energy needed to ionize the hydrogen atom.
We can find the wavelength of a spectral line.
We can find the energy change of the electron moving between two levels.
The first atomic model to adequately explain the radiation spectra of atomic hydrogen was Bohr's model of the hydrogen atom. The atomic Hydrogen model was first presented by Niels Bohr in 1913. Rutherford's model of the hydrogen atom leaves several holes, which Bohr's model of the hydrogen atom tries to fill in.
It has a particular place in history since it introduced the quantum theory, which led to the development of quantum mechanics. Bohr proposed that electrons moved in predetermined orbits or shells with defined radii around the nucleus. It was impossible for electrons to reside between any shells other than those having a radius given by the equation below.
Learn more about hydrogen atom here;
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This question is incomplete, the complete question is;
Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).
how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers
Answer:
Given the data in question;
Dipole moment P = 1 × 10⁻⁹ C.m
now dipole pointing to the right;
P→
(-) ---------------->(+)
d
so let distance between the dipoles be d
∴ P = d
Let = 1 nC
so
P = d
1 × 10⁻⁹ = 1 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (1 × 10⁻⁹)
d = 1 m
Also Let = 2 nC
so
P = d
1 × 10⁻⁹ = 2 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (2 × 10⁻⁹)
d = 0.5 m
Also Let = 3 nC
so
P = d
1 × 10⁻⁹ = 3 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (3 × 10⁻⁹)
d = 0.33 m
such that;
charge distance
1 nC 1.00 m
2 nC 0.50 m
3 nc 0.33 m
4 nC 0.25 m
5 nC 0.20 m